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M8
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Professor
M8
From 8 computers different price, 3 will be selected. What is the probability that 2 most expensive computers will be selected?
= 6/8c3 = 6/56 = 3/28


3/28 is my choice too. 6 = 6C1.

- Vipin
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Professor
M8
From 8 computers different price, 3 will be selected. What is the probability that 2 most expensive computers will be selected?
= 6/8c3 = 6/56 = 3/28


Professor what is 6 in numerator, the number of not most expensive computers? And why do you use this number.
Thank you.
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itishaj
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M8
Professor
M8
From 8 computers different price, 3 will be selected. What is the probability that 2 most expensive computers will be selected?
= 6/8c3 = 6/56 = 3/28

Professor what is 6 in numerator, the number of not most expensive computers? And why do you use this number.
Thank you.


Numerator is 2C2 * 6C1

2C2 ..ways to select 2 expensive ones from 2 computers.6C1...select any other one computer.

thus numerator is 6.
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Professor
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M8,

itishaj explained well. follow this posts.

itishaj
M8
Professor
M8
From 8 computers different price, 3 will be selected. What is the probability that 2 most expensive computers will be selected?
= 6/8c3 = 6/56 = 3/28

Professor what is 6 in numerator, the number of not most expensive computers? And why do you use this number.
Thank you.

Numerator is 2C2 * 6C1

2C2 ..ways to select 2 expensive ones from 2 computers.6C1...select any other one computer.

thus numerator is 6.
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shampoo
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i dont understand this problem, can some explain me in english first? I think i need a lesson on probability, any good resource? thanks
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MA466
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Hi,
Can anyone explain how you got 3/14? Thanks.

ma466
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MA466
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pls explain this problem. How it is 3/14?
ma466



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